Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}2x-8y &= -8 \\ 2x+4y &= -5\end{align*}$
Explanation: Begin by moving the $x$ -term in the second equation to the right side of the equation. $4y = -2x-5$ Divide both sides by $4$ to isolate $y$ $y = {-\dfrac{1}{2}x - \dfrac{5}{4}}$ Substitute this expression for $y$ in the first equation. $2x-8({-\dfrac{1}{2}x - \dfrac{5}{4}}) = -8$ $2x + 4x + 10 = -8$ Simplify by combining terms, then solve for $x$ $6x + 10 = -8$ $6x = -18$ $x = -3$ Substitute $-3$ for $x$ back into the top equation. $2( -3)-8y = -8$ $-6-8y = -8$ $-8y = -2$ $y = \dfrac{1}{4}$ The solution is $\enspace x = -3, \enspace y = \dfrac{1}{4}$.